Calculating Dependant Probability of Multiple Events

There are $\binom{5850}{8}$ ways to choose the winning tickets. The number of combinations in which you have exactly $k$ winning tickets is $\binom{290}{k} \binom{5560}{8-k}$ (choose $k$ winning tickets among yours and $8-k$ amont the remaining $5560$. So the probability is \[ \frac{\binom{290}{k} \binom{5560}{8-k}}{\binom{5850}{8}} = \frac{290! \cdot 5560! \cdot 5842! \cdot 8!}{(290-k)! \cdot k! \cdot (5552+k)! \cdot (8-k)! \cdot 5850!}. \]

For example, for $k=1$ you get \[ \frac{290! \cdot 5560! \cdot 5842! \cdot 8!}{289! \cdot 1! \cdot 5553! \cdot 7! \cdot 5850!} = \frac{290 \cdot 8 \cdot 5560 \cdot 5559 \cdot 5558 \cdot 5557 \cdot 5556 \cdot 5555 \cdot 5554}{5850 \cdot 5849 \cdot 5848 \cdot 5847 \cdot 5846 \cdot 5845 \cdot 5844 \cdot 5843} \simeq 27.8\%. \] Notice that it's not the same as what you did, and actually your formula is wrong: for example, if there were $1000$ winning tickets instead of $8$, your calculation would give $290 \cdot 1000 / 5850 \simeq 4957\%$, which of course doesn't make any sense.

For different $k$, just substitute in the formula above, and remember to simplify the factorials before putting that into a calculator.